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The polynomial $p(x)=x^4+x+1$ can be shown to be irreducible over $\mathbb{Z}_7$. Show that $\mathbb{Z}_7[x]/\langle p(x)\rangle$ is a field.

Since $p(x)$ is irreducible over $\mathbb{Z}_7$, then $\mathbb{Z}_7[x]/\langle p(x)\rangle$ has $7^4$ elements. I know this guarantees the existence of a finite field over $\mathbb{Z}_7$. But I don't know how to relate what I know to the conclusion that $\mathbb{Z}_7[x]/\langle p(x)\rangle$ is a field.

Any help/hints would be appreciated. ^_^

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    $\begingroup$ You already know (presumably) that it's a ring. To show that it's a field, you just need to show that any nonzero element has an inverse. Do you know the Euclidean algorithm? This is where the irreducibility of $p(x)$ comes into play - if $(f(x), p(x)) \neq 1$ then you know that $f(x)$ is a multiple of $p(x)$. $\endgroup$ – Steven Stadnicki Oct 11 '13 at 3:18
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    $\begingroup$ Note: $\langle p(x)\rangle$ is not an ideal of $\Bbb{Z}_7$ in any way. Undoubtedly you are asking about $\Bbb{Z}_7[x]/\langle p(x)\rangle$. What do you know about those ideals of the polynomial ring over a field that are generated by an irreducible polynomial? $\endgroup$ – Jyrki Lahtonen Oct 11 '13 at 3:31
  • $\begingroup$ Why is $\langle p(x)\rangle$ not an ideal of $\mathbb{Z}_7$? I don't know much about ideals. $\endgroup$ – Desperate Fluffy Oct 11 '13 at 3:33
  • $\begingroup$ @DesperateFluffy because it's not even a member of $\mathbb{Z}_7$! The latter is, of course, the set $\{0, 1, 2, \ldots, 6\}$; you mean the ring of polynomials over $\mathbb{Z}_7$. $\endgroup$ – Steven Stadnicki Oct 11 '13 at 3:55
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Hint: As the comments indicate, you want to show that $p(X)$ is irreducible. So suppose not, then $p(X) = f(X)g(X)$ for some $f, g\in \mathbb{Z}_7[X]$, then we may assume without loss of generality that $f(X)$ and $g(X)$ are both monic (why?). Now,

  1. If either $f(X)$ or $g(X)$ has degree 1, then $p(X)$ must have a root on $\mathbb{Z}_7$. Check that this is impossible.
  2. If $f(X) = x^2 + ax+b$ and $g(X) = x^2+cx + d$, you can compare coefficients to get $$ c+a = 0; \qquad ac+d+b = 0 $$ $$ ad+bc = 1; \qquad bd = 1 $$ Can you solve these to get some sort of contradiction? (Remember, you are living over a field, so things should get easier).

Added : Once you know $p(X)$ is irreducible, then you want to show that for any $f(X) \in \mathbb{Z}_7[X]$. Try modifying this solution to suit your needs.

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    $\begingroup$ I'm giving that $p(x)$ is irreducible over $\mathbb{Z}_7$. I want to show that $\mathbb{Z}_7/\langle p(x)\rangle$ is a field. $\endgroup$ – Desperate Fluffy Oct 11 '13 at 4:16
  • $\begingroup$ Thank you for the link! ^_^ $\endgroup$ – Desperate Fluffy Oct 11 '13 at 5:05
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I think this is easiest to prove in its full generality; the specifics of ${\mathbb Z}_7$ and $p(x) = x^4 + x + 1$ play no role at all and only distract.

Theorem. Let $K$ be a field and let $p(x) \in K[x]$ be irreducible. Then $K[x]/(p(x))$ is a field.

Proof. (assuming that you already know that $K[x]/(p(x))$ is a commutative ring). We have to show that every non-zero element of $K[x]/(p(x))$ has an inverse.

Take $0 \neq a \in K[x]$. Then $a = \overline{q(x)}$ for some $q(x) \in K[x]$ and we can take $q(x)$ to be of degree less than $\deg(p(x))$. Because $p(x)$ is irreducible (and because $p(x) \not\mid q(x)$ because of the degree), $p(x)$ and $q(x)$ are relatively prime. Hence, there are $f(x), g(x) \in K[x]$ such that $$f(x) p(x) + g(x) q(x) = 1.$$ In $K[x]/(p(x))$, this equation becomes $$\overline{g(x)} \; \overline{q(x)} = 1,$$ so $\overline{g(x)}$ is the inverse of $\overline{q(x)} = a$.

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